# Derivative of Field-Operator and Vector-Potential

1. Jan 15, 2014

### Abigale

Hello,

I regard field-operators, whereby $\Psi$ is a fermionic-annihilation-operator and

$\vec{A(\vec{r})}$ is an electromagnetic-vector-potential.

Is it possible to do the following step?

$$\nabla \vec{A} \Psi =\Psi \nabla \vec{A} + \vec{A}\nabla\Psi$$

And if its correct, why?

Thx
Abby

2. Jan 15, 2014

### Abigale

Ok I think I got a Solution.
If I consider Coulomb Gauge

$$\nabla \vec{A}=0$$
and I can write
$$\nabla \vec{A} \Psi = \vec{A}\nabla\Psi$$

3. Jan 17, 2014

### Avodyne

$$\nabla\cdot( \vec{A} \Psi) =(\nabla\cdot \vec{A})\Psi + \vec{A}\cdot\nabla\Psi$$
by the product rule for derivatives. If you like, you can move $\Psi$ to the left in the first term, since $\Psi$ and $\vec A$ commute.